LECTURE 16: SOLVING THE LAPLACE EQUATIONS
MINGFENG ZHAO
July 31, 2015
Laplace equation
The Laplace equation (i.e., ∆u = 0) arises as a steady state problem for the Heat or Wave equation that do not vary
with time, that is, ut = utt = 0.
2D:
∆u =
∂2u ∂2u
+ 2 = 0.
∂x2
∂y
3D:
∆u =
∂2u ∂2u ∂2u
+ 2 + 2 = 0.
∂x2
∂y
∂z
For Laplace equation, ∆u = 0, we do NOT have initial conditions, and ONLY have boundary conditions. For the
boundary condition, there are three common types: Let D be a domain in R2 , and ∂D denote the boundary of D:
• Dirichlet boundary condition:
BC:
u(x, y) = f (x, y),
(x, y) ∈ ∂D.
• Neumann boundary condition:
BC:
∂u
(x, y) = f (x, y),
∂n
(x, y) ∈ ∂D.
• Robin boundary condition:
BC:
a(x, y)u(x, y) + b(x, y)
∂u
(x, y) = f (x, y),
∂n
(x, y) ∈ ∂D.
Remark 1. For any (x, y) ∈ ∂D, we have a unit normal vector n(x, y) along the boundary of D at point (x, y), then
∂u
= ∇u(x, y) · n(x, y).
∂n
1
2
MINGFENG ZHAO
Laplace equation in two dimensions
In this course, we consider the domains D that are rectangular, circular and pizza slices.
Remark 2. When the domain D is either circular or pizza slice, we are going to use the polar coordinates form of
the Laplacian:
∆u =
∂ 2 u 1 ∂u
1 ∂2u
+
= 0.
+
∂r2
r ∂r
r2 ∂θ2
Recall the polar coordinates:
x = r cos(θ),
y = r sin(θ).
Then
∂
∂r
∂
∂θ
∂ ∂x
∂ ∂y
·
+
·
∂x ∂r
∂y ∂r
∂
∂
+ sin(θ)
= cos(θ)
∂x
∂y
∂ ∂x
∂ ∂y
=
·
+
·
∂x ∂θ
∂y ∂θ
∂
∂
+ r cos(θ) .
= −r sin(θ)
∂x
∂y
=
Then we have
∂
∂
sin(θ) ∂
= cos(θ)
−
,
∂x
∂r
r ∂θ
and
∂
∂
cos(θ) ∂
= sin(θ)
+
.
∂y
∂r
r ∂θ
That is, we have
rx = cos(θ),
ry = sin(θ),
θx = −
sin(θ)
,
r
and θy =
cos(θ)
.
r
Then
ux
uy
=
ur rx + uθ θx
=
cos(θ)ur −
=
ur ry + uθ θy
=
sin(θ)ur +
sin(θ)
uθ
r
cos(θ)
uθ
r
uxx
cos(θ)
sin(θ)
sin(θ)
θx −
r
[urθ rx + uθθ θx ]
x uθ −
2
r
r
r
sin(θ)
sin(θ)
cos(θ) sin(θ) sin(θ)
= sin(θ) ·
ur + cos(θ) cos(θ)urr −
urθ − −
·
−
· cos(θ) uθ
r
r
r
r
r2
=
− sin(θ)θx ur + cos(θ)[urr rx + urθ θx ] −
LECTURE 16: SOLVING THE LAPLACE EQUATIONS
sin(θ)
sin(θ)
−
cos(θ)urθ −
uθθ
r
r
=
sin(θ) cos(θ)
sin(θ) cos(θ) sin(θ) cos(θ)
sin2 (θ)
2
+
uθ
ur + cos (θ)urr −
urθ +
r
r
r2
r2
sin(θ) cos(θ)
sin2 (θ)
uθθ
urθ +
r
r2
sin2 (θ)
sin2 (θ)
2 sin(θ) cos(θ)
2 sin(θ) cos(θ)
uθ +
uθθ
cos2 (θ)urr +
ur −
urθ +
2
r
r
r
r2
cos(θ)
sin(θ)
cos(θ)
r
cos(θ)θy ur + sin(θ)[urr ry + urθ θy ] + −
θy −
[urθ ry + uθθ θy ]
y uθ +
2
r
r
r
cos(θ)
cos(θ)
sin(θ) cos(θ) cos(θ)
· sin(θ) uθ
cos(θ) ·
ur + sin(θ) sin(θ)urr +
urθ + −
·
−
r
r
r
r
r2
cos(θ)
cos(θ)
+
sin(θ)urθ +
uθθ
r
r
2
sin(θ) cos(θ)
sin(θ) cos(θ) sin(θ) cos(θ)
cos (θ)
2
ur + sin (θ)urr +
urθ + −
−
uθ
r
r
r2
r2
−
=
uyy
=
=
=
cos2 (θ)
sin(θ) cos(θ)
urθ +
uθθ
r
r2
cos2 (θ)
2 sin(θ) cos(θ)
2 sin(θ) cos(θ)
cos2 (θ)
= sin2 (θ)urr +
ur +
urθ −
uθ +
uθθ
2
r
r
r
r2
+
Hence we have
1
1
∆u = uxx + uyy = urr + ur + 2 uθθ .
r
r
Example 1. Let u(x, y) = ln |(x, y)| = ln r, then
ur =
1
1
, urr = − 2 and uθ = uθθ = 0.
r
r
So we get
∆u = −
The function −
1 1
1
+ · + 0 = 0.
r2
r r
1
ln |(x, y)| is called the fundamental solution of ∆u = 0 in the two dimensional space.
2π
Solving the Laplace equation with Dirichlet boundary condition in a rectangle
For the Laplace equation in a rectangle D = [0, a] × [0, b], since ∂D consists of four line segments, then



∆u = uxx + uyy = 0, (x, y) ∈ D,



(1)
BC : u(x, 0) = f1 (x), u(x, b) = f2 (x), 0 ≤ x ≤ a,





u(0, y) = g1 (y), u(a, y) = g2 (y), 0 ≤ y ≤ b.
3
4
MINGFENG ZHAO
To solve (1), we need to solve the following four problems:



∆u = uxx + uyy = 0, (x, y) ∈ D,



(2)
BC : u(x, 0) = f1 (x), u(x, b) = 0, 0 ≤ x ≤ a,





u(0, y) = 0, u(a, y) = 0, 0 ≤ y ≤ b.



∆u = uxx + uyy = 0, (x, y) ∈ D,



(3)
BC : u(x, 0) = 0, u(x, b) = f2 (x), 0 ≤ x ≤ a,





u(0, y) = 0, u(a, y) = 0, 0 ≤ y ≤ b.



∆u = uxx + uyy = 0, (x, y) ∈ D,



(4)
BC : u(x, 0) = 0, u(x, b) = 0, 0 ≤ x ≤ a,





u(0, y) = g1 (y), u(a, y) = 0, 0 ≤ y ≤ b.



∆u = uxx + uyy = 0, (x, y) ∈ D,



(5)
BC : u(x, 0) = 0, u(x, b) = 0, 0 ≤ x ≤ a,





u(0, y) = 0, u(a, y) = g2 (y), 0 ≤ y ≤ b.
It’s easy to see that the solution to (1) is just the sum of solutions to (2), (3), (4), and (5). We are going to solve (2),
(3), (4), and (5) by using the method of separation of variables.
In the following, we just solve (2). First, let’s look the problem:



∆u = uxx + uyy = 0, (x, y) ∈ D,



(6)
BC : u(x, b) = 0, 0 ≤ x ≤ a,





u(0, y) = 0, u(a, y) = 0, 0 ≤ y ≤ b.
Let u(x, y) = X(x)Y (y) be a non-zero separated solution to (6), then X 00 (x)Y (y) + X(x)Y 00 (y) = 0, that is,
−
Y 00 (y)
X 00 (x)
=
= λ.
X(x)
Y (y)
It’s easy to see that λx = λy = 0, that is, λ is a constant. Since u(0, y) = u(a, y) = 0, then X(0) = X(a) = 0. Then
X(x) satisfies:
(7)

 X 00 + λX = 0,
 X(0) = X(a) = 0.
For the eigenvalue problem (7), we know that the eigenvalues are λ =
nπ are X(x) = C sin
x , with n ∈ N.
a
nπ 2
a
, and the corresponding eigenfunctions
LECTURE 16: SOLVING THE LAPLACE EQUATIONS
For λ =
nπ 2
(8)
a
5
, since u(x, b) = 0, then Y (b) = 0. Then Y satisfies

2

 Y 00 − λY = Y 00 − nπ Y = 0,
a

 Y (b) = 0.
Then we know that
i
h nπ
2nπ
nπ
Y (y) = C e a y − e a b e− a y .
Then for any n ∈ N, we can find a non-zero solution to (7):
h nπ
i
nπ 2nπ
nπ
un (x, y) = e a y − e a b e− a y sin
x .
a
Let
u(x, y) =
∞
X
h nπ
i
nπ 2nπ
nπ
an e a y − e a b e− a y sin
x
a
n=1
by the solution to (3), it’s easy to see that u(x, b) = 0 for 0 ≤ x ≤ a, and u(0, y) = u(a, y) = 0 for 0 ≤ y ≤ b. Since
u(x, 0) = f1 (x) for 0 ≤ x ≤ a, then
f1 (x) =
∞
X
h
i
nπ 2nπ
an 1 − e a b sin
x .
a
n=1
So we get
i 2Z a
nπ h
2nπ
b
a
f1 (x) sin
x dx,
=
an 1 − e
a 0
a
for all n ≥ 1.
So we get
2
i
an = h
2nπ
a 1−e a b
Z
a
f1 (x) sin
0
nπ x dx,
a
for all n ≥ 1.
Therefore, the solution to (3) is given by:
u(x, y) =
∞
X
h nπ
i
nπ 2nπ
nπ
an e a y − e a b e− a y sin
x ,
a
n=1
where
an =
2
h
i
2nπ
a 1−e a b
Z
a
f1 (x) sin
0
nπ x dx,
a
for all n ≥ 1.
Remark 3. Here are some useful functions:
Hyperbolic sine :
Hyperbolic cosine :
Hyperbolic tangent :
ex − e−x
2
ex + e−x
cosh(x) =
2
sinh(x)
ex − e−x
= x
.
tanh(x) =
cosh(x)
e + e−x
sinh(x) =
6
MINGFENG ZHAO
Solving the Laplace equation with Dirichlet boundary condition in a semi-infinite strip
Consider the following problem:



∆u = uxx + uyy = 0, 0 < x < a, y > 0,






 BC : u(x, 0) = f (x), 0 ≤ x ≤ a,
(9)



u(a, y) = u(0, y) = 0, y ≥ 0,






lim u(x, y) = 0, x > 0.
y→∞
First, let’s look the problem:
(10)











∆u = uxx + uyy = 0,
BC : u(0, y) = u(a, y) = 0,
lim u(x, y) = 0,
y→∞
(x, y) ∈ D,
y ≥ 0,
x > 0.
Let u(x, y) = X(x)Y (y) be a non-zero separated solution to (10), then X 00 (x)Y (y) + X(x)Y 00 (y) = 0, that is,
−
X 00 (x)
Y 00 (y)
=
= λ.
X(x)
Y (y)
It’s easy to see that λx = λy = 0, that is, λ is a constant. Since u(0, y) = u(a, y) = 0, then X(0) = X(a) = 0. Then
X(x) satisfies:
(11)

 X 00 + λX = 0,
 X(0) = X(a) = 0.
LECTURE 16: SOLVING THE LAPLACE EQUATIONS
7
nπ 2
For the eigenvalue problem (11), we know that the eigenvalues are λ =
, and the corresponding eigenfunctions
a
nπ are X(x) = C sin
x , with n ∈ N.
nπ 2 a
nπ 2
For λ =
, since u(x, b) = 0, then Y (b) = 0. Then Y satisfies Y 00 − λY = Y 00 −
Y = 0, then
a
a
Y (y) = C1 e
nπ
a y
+ C2 e−
nπ
a
Y.
Since we need lim u(x, y) = 0 for x > 0, then we must have C1 = 0. Hence for any n ∈ N, we can find a non-zero
y→∞
solution to (10):
un (x, y) = e−
nπ
a y
nπ x .
a
sin
Let
u(x, y) =
∞
X
an e−
nπ
a y
sin
n=1
nπ x
a
be the solution to (9), since u(x, 0) = f (x) for 0 ≤ x ≤ a, then
f (x) =
∞
X
an sin
n=1
nπ x .
a
Then
an =
a
Z
2
a
f (x) sin
0
nπ x ,
a
for all n ≥ 1.
Therefore, the solution to (9) is given by:
u(x, y) =
∞
X
an e−
nπ
a y
sin
n=1
nπ x ,
a
where
2
an =
a
Z
a
f (x) sin
0
nπ x dx,
a
for all n ≥ 1.
Remark 4. For the Laplace equation on a unbounded domain, it’s usual to put extra conditions at infinity to make
1
sure that the problem will have only one solution. For example, let u(x, y) = sin(x)y + sin(x)y 3 , then u satisfies:
6



∆u = uxx + uyy = 0, 0 < x < π, y > 0,




BC : u(x, 0) = 0, 0 ≤ x ≤ π,






u(π, y) = u(0, y) = 0, y ≥ 0.
Department of Mathematics, The University of British Columbia, Room 121, 1984 Mathematics Road, Vancouver, B.C.
Canada V6T 1Z2
E-mail address: mingfeng@math.ubc.ca